Proof of the Markov property of BM using intersection stable generators.

Question

I am looking at the following proof of the Markov property of BM. There are two points that I'm not clear about. First, what does it mean precisely that $\bigcup_{0<s_1<\cdots <s_m\le a,\;m\ge 1}\sigma(B(s_j):j=1,\dots ,m)$ and $\bigcup_{0<u_1<\cdots <u_n,\;n\ge 1}\sigma(B(u_k):k=1,\dots ,n)$ are independent? I'm not sure what two unions of sigma-algebras being independent means. Does it mean that if we take any sequence of points from either side, then the sigma algebras generated by $B(s_j)$s and $W(u_k)$s are independent? This is the way I interpreted it and so since we have established $\sigma(B(s_j): j=1,\dots , m) \perp \sigma(W(u_k):k=1,\dots,n)$ for any sequence of points $s_j$ and $u_k$, then we get for any $u_k$s, $\sigma(W(u_k):k=1,\dots,n) \perp \bigcup_{0<s_1<\cdots <s_m\le a,\;m\ge 1}\sigma(B(s_j):j=1,\dots ,m)$ and since $u_k$s are arbitrary, we get the result. Is this the correct interpretation?

Second, why are these families $\cap$-stable? This means that $\sigma(B(s_j):j=1,\dots,m)\cap \sigma(B(s_k):k=1, \dots, n)$ also belongs to this union. But do we have $\sigma(B(s_j):j=1,\dots,m)\cap \sigma(B(s_k):k=1, \dots, n)=\sigma(B(s_t): t=j \;\text{and}\; k)$? I think this is the only way to see that the family is $\cap-$stable, and I can see that $\supset$ holds but I don't know how to prove the other direction. I would greatly appreciate any help.

Answer 1

Two families of sets $\mathcal{G}$ and $\mathcal{H}$ are independent if, and only if,

$$\forall G \in \mathcal{G}, H \in \mathcal{H}: \, \, \mathbb{P}(G \cap H) = \mathbb{P}(G) \mathbb{P}(H).$$

Note that $\mathcal{G}$ and $\mathcal{H}$ do not need to be $\sigma$-algebras; in your framework we have

$$\begin{align*} \mathcal{G} &:= \bigcup_{m \geq 1} \bigcup_{0<s_1<\ldots<s_m \leq a} \sigma(B(s_1),\ldots,B(s_m)) \\ \mathcal{H} &:= \bigcup_{n \geq 1} \bigcup_{0<u_1 < \ldots<u_n} \sigma(B(u_1),\ldots,B(u_n)). \end{align*}$$

Why are these families $\cap$-stable?

Let $G_1, G_2 \in \mathcal{G}$, then we can find $0<s_1<\ldots<s_m \leq a$ and $0<t_1<\ldots<t_k \leq a$ such that $G_1 \in \sigma(B(s_j);j=1,\ldots,m)$ and $G_2 \in \sigma(B(t_j); j=1,\ldots,k)$. Now the idea is simply to "merge" the two sequences $(s_j)_{j=1,\ldots,m}$ and $(t_j)_{j=1,\ldots,k}$: Clearly, there exist $\ell \leq k+m$ and $0<r_1<\ldots<r_{\ell} \leq a$ such that

$$\{s_j;j=1,\ldots,m\} \cup \{t_j;j=1,\ldots,k\} = \{r_j; j=1,\ldots,\ell\}.$$

Then

$$G_1 \in \sigma(B(s_j);j=1,\ldots,m) \subseteq \sigma(B(r_j);j=1,\ldots,\ell)$$

and

$$G_2 \in \sigma(B(t_j);j=1,\ldots,k) \subseteq \sigma(B(r_j);j=1,\ldots,\ell)$$

and so

$$G_1 \cap G_2 \in \sigma(B(r_j);j=1,\ldots,\ell) \subseteq \mathcal{G}.$$

A very similar reasoning shows that $\mathcal{H}$ is $\cap$-stable.